Question: The following equation is true for all real values of $k$ for which the expression on the left is defined, and $D$ is a polynomial expression. $\dfrac{3k^2+24k}{6k^2+12k} \div \dfrac{D}{2k^2-12k-32}=1$ What is $D$ ? $D=$
Answer: The left side of the equation is a quotient of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting quotient on the left side should cancel out completely. In order to solve for $D$, let's divide the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $3k^2+24k$, of the dividend can be factored as $3k(k+8)$ by factoring out $3k$. The denominator, $6k^2+12k$, of the dividend can be factored as $6k(k+2)$ by factoring out a $6k$. The denominator, $2k^2-12k-32$, of the divisor can be factored as $2(k-8)(k+2)$ by factoring out $2$ and using the sum-product pattern. Now the quotient looks as follows: $\dfrac{3k(k+8)}{6k(k+2)} \div \dfrac{D}{2(k-8)(k+2)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{3k(k+8)}{6k(k+2)} \div \dfrac{D}{2(k-8)(k+2)} \\\\\\ &=\dfrac{3k(k+8)}{6k(k+2)} \cdot \dfrac{2(k-8)(k+2)}{D} &\text{Flip the divisor} \\\\\\ &= \dfrac{3k(k+8) \cdot 2(k-8)(k+2)}{6k(k+2) \cdot D} &\text{Multiply across.}\\\\\\ &= \dfrac{{\cancel{3}}{\cancel{k}}(k+8) {\cancel{2}}(k-8){\cancel{(k+2)}}}{{\cancel{3}}\cdot {\cancel{2}}{\cancel{k}}{\cancel{(k+2)}} D} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(k+8)(k-8)}{D} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{(k+8)(k-8)}{D}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $D=(k+8)(k-8)$, which is equivalent to $k^2-64$.